3x^2+65x-575=x^2+20x-125

Solution for 3x^2+65x-575=x^2+20x-125 equation:

3x^2+65x-575=x^2+20x-125

We move all terms to the left:

3x^2+65x-575-(x^2+20x-125)=0

We get rid of parentheses

3x^2-x^2+65x-20x+125-575=0

We add all the numbers together, and all the variables

2x^2+45x-450=0

a = 2; b = 45; c = -450;
Δ = b2-4ac
Δ = 452-4·2·(-450)
Δ = 5625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5625}=75$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-75}{2*2}=\frac{-120}{4}
=-30
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+75}{2*2}=\frac{30}{4}
=7+1/2
$