7(x^2+2x)=3x^2+22x

Solution for 7(x^2+2x)=3x^2+22x equation:

7(x^2+2x)=3x^2+22x

We move all terms to the left:

7(x^2+2x)-(3x^2+22x)=0

We multiply parentheses

7x^2+14x-(3x^2+22x)=0

We get rid of parentheses

7x^2-3x^2+14x-22x=0

We add all the numbers together, and all the variables

4x^2-8x=0

a = 4; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·4·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*4}=\frac{0}{8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*4}=\frac{16}{8}
=2
$