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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/qw1zd3pddub5avi5q4j30byom1vx821mtg.png)
the solutions are
![\begin{gathered} x1=\frac{11+_{}\sqrt[\square]{481}}{4} \\ \text{and} \\ x2=\frac{11-\sqrt[\square]{481}}{4} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/knh611q9pkyeahshhi288g7p6ysd127uhp.png)
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=(-13\pm3)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/1qvt0wsoskw0hmk88r9dlqhhczehqcfi8c.png)
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=(6\pm12)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/bkjrb6j7uwzdgj5kxolbgzmg0hjxm1yxli.png)
so
x=9 and x=-3