20x^2+25x=26x^2+4x-27

Solution for 20x^2+25x=26x^2+4x-27 equation:

20x^2+25x=26x^2+4x-27

We move all terms to the left:

20x^2+25x-(26x^2+4x-27)=0

We get rid of parentheses

20x^2-26x^2+25x-4x+27=0

We add all the numbers together, and all the variables

-6x^2+21x+27=0

a = -6; b = 21; c = +27;
Δ = b2-4ac
Δ = 212-4·(-6)·27
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-33}{2*-6}=\frac{-54}{-12}
=4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+33}{2*-6}=\frac{12}{-12}
=-1
$