(-x+5)=9x^2+9x

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

(-x+5)=9x^2+9x

We move all terms to the left:

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

We add all the numbers together, and all the variables

(-1x+5)-(9x^2+9x)=0

We get rid of parentheses

-9x^2-1x-9x+5=0

We add all the numbers together, and all the variables

-9x^2-10x+5=0

a = -9; b = -10; c = +5;
Δ = b2-4ac
Δ = -102-4·(-9)·5
Δ = 280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{70}}{2*-9}=\frac{10-2\sqrt{70}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{70}}{2*-9}=\frac{10+2\sqrt{70}}{-18}
$