-4x2-32x-50=-x2-3+6

Solution for -4x2-32x-50=-x2-3+6 equation:

-4x^2-32x-50=-x2-3+6

We move all terms to the left:

-4x^2-32x-50-(-x2-3+6)=0

We add all the numbers together, and all the variables

-4x^2-(-1x^2-3+6)-32x-50=0

We get rid of parentheses

-4x^2+1x^2-32x+3-6-50=0

We add all the numbers together, and all the variables

-3x^2-32x-53=0

a = -3; b = -32; c = -53;
Δ = b2-4ac
Δ = -322-4·(-3)·(-53)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{97}}{2*-3}=\frac{32-2\sqrt{97}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{97}}{2*-3}=\frac{32+2\sqrt{97}}{-6}
$