4x^2+7-40=0

Solution for 4x^2+7-40=0 equation:

4x^2+7-40=0

We add all the numbers together, and all the variables

4x^2-33=0

a = 4; b = 0; c = -33;
Δ = b2-4ac
Δ = 02-4·4·(-33)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{33}}{2*4}=\frac{0-4\sqrt{33}}{8}
=-\frac{4\sqrt{33}}{8}
=-\frac{\sqrt{33}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{33}}{2*4}=\frac{0+4\sqrt{33}}{8}
=\frac{4\sqrt{33}}{8}
=\frac{\sqrt{33}}{2}
$