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

Solution for -9-9x=-8x^2-5+x equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

8x^2-10x-4=0

a = 8; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·8·(-4)
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{57}}{2*8}=\frac{10-2\sqrt{57}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{57}}{2*8}=\frac{10+2\sqrt{57}}{16}
$