0=2x^2+4x-7

Solution for 0=2x^2+4x-7 equation:

0=2x^2+4x-7

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-2x^2-4x+7=0

a = -2; b = -4; c = +7;
Δ = b2-4ac
Δ = -42-4·(-2)·7
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{-(-4)-6\sqrt{2}}{2*-2}=\frac{4-6\sqrt{2}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{2}}{2*-2}=\frac{4+6\sqrt{2}}{-4}
$