3x(x-3)+22=2x^2+4(x-5)

Solution for 3x(x-3)+22=2x^2+4(x-5) equation:

3x(x-3)+22=2x^2+4(x-5)

We move all terms to the left:

3x(x-3)+22-(2x^2+4(x-5))=0

We multiply parentheses

3x^2-9x-(2x^2+4(x-5))+22=0


We calculate terms in parentheses: -(2x^2+4(x-5)), so:

2x^2+4(x-5)

We multiply parentheses

2x^2+4x-20

Back to the equation:

-(2x^2+4x-20)


We get rid of parentheses

3x^2-2x^2-9x-4x+20+22=0

We add all the numbers together, and all the variables

x^2-13x+42=0

a = 1; b = -13; c = +42;
Δ = b2-4ac
Δ = -132-4·1·42
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-1}{2*1}=\frac{12}{2}
=6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*1}=\frac{14}{2}
=7
$