10x-40=(3x^2)-90

Solution for 10x-40=(3x^2)-90 equation:

10x-40=(3x^2)-90

We move all terms to the left:

10x-40-((3x^2)-90)=0

We get rid of parentheses

-3x^2+10x+90-40=0

We add all the numbers together, and all the variables

-3x^2+10x+50=0

a = -3; b = 10; c = +50;
Δ = b2-4ac
Δ = 102-4·(-3)·50
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{7}}{2*-3}=\frac{-10-10\sqrt{7}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{7}}{2*-3}=\frac{-10+10\sqrt{7}}{-6}
$