X^2+7x-5=x+4

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

X^2+7X-5=X+4

We move all terms to the left:

X^2+7X-5-(X+4)=0

We get rid of parentheses

X^2+7X-X-4-5=0

We add all the numbers together, and all the variables

X^2+6X-9=0

a = 1; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·1·(-9)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{2}}{2*1}=\frac{-6-6\sqrt{2}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{2}}{2*1}=\frac{-6+6\sqrt{2}}{2}
$