4x^2+40x-21=7

Solution for 4x^2+40x-21=7 equation:

4x^2+40x-21=7

We move all terms to the left:

4x^2+40x-21-(7)=0

We add all the numbers together, and all the variables

4x^2+40x-28=0

a = 4; b = 40; c = -28;
Δ = b2-4ac
Δ = 402-4·4·(-28)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-32\sqrt{2}}{2*4}=\frac{-40-32\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+32\sqrt{2}}{2*4}=\frac{-40+32\sqrt{2}}{8}
$