3x-4=4x^2-40

Solution for 3x-4=4x^2-40 equation:

3x-4=4x^2-40

We move all terms to the left:

3x-4-(4x^2-40)=0

We get rid of parentheses

-4x^2+3x+40-4=0

We add all the numbers together, and all the variables

-4x^2+3x+36=0

a = -4; b = 3; c = +36;
Δ = b2-4ac
Δ = 32-4·(-4)·36
Δ = 585
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:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{65}}{2*-4}=\frac{-3-3\sqrt{65}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{65}}{2*-4}=\frac{-3+3\sqrt{65}}{-8}
$