x^2+7x-3=x+4+6

Solution for x^2+7x-3=x+4+6 equation:

x^2+7x-3=x+4+6

We move all terms to the left:

x^2+7x-3-(x+4+6)=0

We add all the numbers together, and all the variables

x^2+7x-(x+10)-3=0

We get rid of parentheses

x^2+7x-x-10-3=0

We add all the numbers together, and all the variables

x^2+6x-13=0

a = 1; b = 6; c = -13;
Δ = b2-4ac
Δ = 62-4·1·(-13)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{22}}{2*1}=\frac{-6-2\sqrt{22}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{22}}{2*1}=\frac{-6+2\sqrt{22}}{2}
$