Question

Solve these 3 equations 1. (x - 8) (2x + 5) = 52. x² + 13x + 40 = 03. x² - 6x - 20 = 7

Answer 1

1. (x - 8) (2x + 5) = 5



2. x² + 13x + 40 = 0



3. x² - 6x - 20 = 7​

Part 1

apply distributive property left side

2x^2+5x-16x-40=5

2x^2-11x-45=0

Apply the formula to solve quadratic equation

we have

a=2

b=-11

c=-45

substitute

$\begin{gathered} x=\frac{11\pm\sqrt[\square]{(-11^2)-4(2)(-45)}}{4} \\ \\ x=\frac{11\pm\sqrt[\square]{481}}{4} \end{gathered}$

the solutions are

$\begin{gathered} x1=\frac{11+_{}\sqrt[\square]{481}}{4} \\ \text{and} \\ x2=\frac{11-\sqrt[\square]{481}}{4} \end{gathered}$

Part 2

x² + 13x + 40 = 0

we have

a=1

b=13

c=40

Apply the formula

$\begin{gathered} x=\frac{-13\pm\sqrt[\square]{13^2-4(1)(40)}}{2} \\ \\ x=\frac{-13\pm\sqrt[\square]{9}}{2} \\ \\ x=\frac{-13\pm3}{2} \end{gathered}$

so

x=-5 and x=-8

Part 3

x² - 6x - 20 = 7​

x² - 6x - 27=0

so

a=1

b=-6

c=-27

Apply the formula

$\begin{gathered} x=\frac{6\pm\sqrt[\square]{(-6^2)-4(1)(-27)}}{2} \\ \\ x=\frac{6\pm\sqrt[\square]{144}}{2} \\ \\ x=\frac{6\pm12}{2} \end{gathered}$

so

x=9 and x=-3