2x^2+21=3x^2+4x

Solution for 2x^2+21=3x^2+4x equation:

2x^2+21=3x^2+4x

We move all terms to the left:

2x^2+21-(3x^2+4x)=0

We get rid of parentheses

2x^2-3x^2-4x+21=0

We add all the numbers together, and all the variables

-1x^2-4x+21=0

a = -1; b = -4; c = +21;
Δ = b2-4ac
Δ = -42-4·(-1)·21
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*-1}=\frac{14}{-2}
=-7
$