25x^2-3x+36=63x

Solution for 25x^2-3x+36=63x equation:

25x^2-3x+36=63x

We move all terms to the left:

25x^2-3x+36-(63x)=0

We add all the numbers together, and all the variables

25x^2-66x+36=0

a = 25; b = -66; c = +36;
Δ = b2-4ac
Δ = -662-4·25·36
Δ = 756
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:

$\sqrt{\Delta}=\sqrt{756}=\sqrt{36*21}=\sqrt{36}*\sqrt{21}=6\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-66)-6\sqrt{21}}{2*25}=\frac{66-6\sqrt{21}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-66)+6\sqrt{21}}{2*25}=\frac{66+6\sqrt{21}}{50}
$