3x^2-19x=-2x^2-7x-7

Solution for 3x^2-19x=-2x^2-7x-7 equation:

3x^2-19x=-2x^2-7x-7

We move all terms to the left:

3x^2-19x-(-2x^2-7x-7)=0

We get rid of parentheses

3x^2+2x^2+7x-19x+7=0

We add all the numbers together, and all the variables

5x^2-12x+7=0

a = 5; b = -12; c = +7;
Δ = b2-4ac
Δ = -122-4·5·7
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2}{2*5}=\frac{10}{10}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2}{2*5}=\frac{14}{10}
=1+2/5
$