2x2+3x-7=x^2+5x+39

Solution for 2x2+3x-7=x^2+5x+39 equation:

2x^2+3x-7=x^2+5x+39

We move all terms to the left:

2x^2+3x-7-(x^2+5x+39)=0

We get rid of parentheses

2x^2-x^2+3x-5x-39-7=0

We add all the numbers together, and all the variables

x^2-2x-46=0

a = 1; b = -2; c = -46;
Δ = b2-4ac
Δ = -22-4·1·(-46)
Δ = 188
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{47}}{2*1}=\frac{2-2\sqrt{47}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{47}}{2*1}=\frac{2+2\sqrt{47}}{2}
$