x^2+x+x-3=4

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

x^2+x+x-3=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+2x-7=0

a = 1; b = 2; c = -7;
Δ = b2-4ac
Δ = 22-4·1·(-7)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{2}}{2*1}=\frac{-2-4\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{2}}{2*1}=\frac{-2+4\sqrt{2}}{2}
$