A Quadratic equation is any equation that has this kind of algebraic format $a{x}^{2}+bx+c=0$

$4{x}^{2}-4x-24=0$

$a{x}^{2}+bx+c=0$

x = $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

a = 4, b = -4, c = -24

Solution:

x = $\frac{-\mathrm{\left(-4\right)}±\sqrt{{\mathrm{-4}}^{2}-4\mathrm{×4}\mathrm{×-24}}}{2\mathrm{×4}}$

x = $\frac{-\mathrm{\left(-4\right)}±\sqrt{\mathrm{16}-\mathrm{\left(-384\right)}}}{8}$

x = $\frac{-\mathrm{\left(-4\right)}±\sqrt{\mathrm{16}+\mathrm{384}}}{8}$

x = $\frac{-\mathrm{\left(-4\right)}±\sqrt{\mathrm{400}}}{8}$

x = $\frac{4±\mathrm{20}}{8}$

x = $\frac{4+\mathrm{20}}{8}$ and x = $\frac{4-\mathrm{20}}{8}$

x = 3, and x = -2

## How to Solve the Quadratic equation example

Solve the equation

$5{x}^{2}-2x-1=0$

a = 5, b = -2, c = -1

Solution:

x = $\frac{-\mathrm{\left(-2\right)}±\sqrt{{\mathrm{\left(-2\right)}}^{2}-4\mathrm{×5}\mathrm{×-1}}}{2\mathrm{×5}}$

x = $\frac{2±\sqrt{4-\left(-20\right)}}{10}$

x = $\frac{2±\sqrt{4+20}}{10}$

x = $\frac{2±\sqrt{\mathrm{24}}}{10}$

x = $\frac{2±\mathrm{4.899}}{10}$

x = $\frac{2+\mathrm{4.899}}{10}$ and $\frac{2-\mathrm{4.899}}{10}$

Answer: x = 0.69 and x = -0.29

Solve the equation $2{x}^{2}+3x-7$ ,giving your answers to 2 decimal places

a = 2

b = 3

c = - 7

Formula:

x = $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

Solution:

$\frac{-3±\sqrt{\left({3}^{2}\right)-\left(4×2×\mathrm{-7}\right)}}{2×2}$

$\frac{-3±\sqrt{\left(9\right)-\left(-56\right)}}{4}$

$\frac{-3±\sqrt{9+56}}{4}$

$\frac{-3±\sqrt{65}}{4}$

x = $\frac{-3+\sqrt{65}}{4}$ or x = $\frac{-3-\sqrt{65}}{4}$

Answer: x = -2.77 or x = 1.27

Find the values of x by Factoring the Quadratic Equation

${x}^{2}+2x-3=0$

Solution:

Product = -3

Sum = 2

Factors = (3, -1)

### Points to note

1. Product = $a×c$

2. Sum = is the addition of factors, from the algebraic format the sum is b

3. Factors = while factors are two numbers which when multiplied they will give us -3 has a product in this case and when added they will give us 2 has the sum in the example above.

${x}^{2}+2x-3=0$

${x}^{2}+\left(3x-x\right)-3=0$

The numbers in the brackets are factors which have replace the sum 2

${x}^{2}+3x-x-3=0$

$x\left(x+3\right)-1\left(x+3\right)=0$

$\left(x-1\right)\left(x+3\right)=0$

$\left(x-1\right)=0$ and $\left(x+3\right)=0$

Answer: x = 1 and x = -3

## Complete the Square

Find the values of x by using Complete the Square of a Quadratic Equation method

${x}^{2}+2x-3=0$

Solution:

${x}^{2}+2x-3=0$

${x}^{2}+2x=3$

### Points to note

1. The coefficient of a = 1, b = 2 and C = -3

2. Using Complete the Square method, the coefficient b is first divided by half ($\frac{1}{2}$) and then power 2 is added to the coefficient.

${x}^{2}+{1}^{2}=3+1$

One (1) is then added to the right hand side of the equation

${\left(\mathrm{x + 1}\right)}^{2}=4$

$\sqrt{{\left(\mathrm{x + 1}\right)}^{2}}=\sqrt{4}$

$\left(\mathrm{x + 1}\right)=±2$

$\left(\mathrm{x + 1}\right)=+2$ and $\left(\mathrm{x + 1}\right)=-2$

$x=+2-1$ and $x=-2-1$

Answer: x = 1 and x = -3