(7x^2+8x-5)+(9x^2-9x)=0

Solution for (7x^2+8x-5)+(9x^2-9x)=0 equation:

(7x^2+8x-5)+(9x^2-9x)=0

We get rid of parentheses

7x^2+9x^2+8x-9x-5=0

We add all the numbers together, and all the variables

16x^2-1x-5=0

a = 16; b = -1; c = -5;
Δ = b2-4ac
Δ = -12-4·16·(-5)
Δ = 321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{321}}{2*16}=\frac{1-\sqrt{321}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{321}}{2*16}=\frac{1+\sqrt{321}}{32}
$