4x+40+x^2+7x=2x^2+5x

Solution for 4x+40+x^2+7x=2x^2+5x equation:

4x+40+x^2+7x=2x^2+5x

We move all terms to the left:

4x+40+x^2+7x-(2x^2+5x)=0

We add all the numbers together, and all the variables

x^2+11x-(2x^2+5x)+40=0

We get rid of parentheses

x^2-2x^2+11x-5x+40=0

We add all the numbers together, and all the variables

-1x^2+6x+40=0

a = -1; b = 6; c = +40;
Δ = b2-4ac
Δ = 62-4·(-1)·40
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*-1}=\frac{8}{-2}
=-4
$