4x^2=1028

Solution for 4x^2=1028 equation:

4x^2=1028

We move all terms to the left:

4x^2-(1028)=0

a = 4; b = 0; c = -1028;
Δ = b2-4ac
Δ = 02-4·4·(-1028)
Δ = 16448
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:

$\sqrt{\Delta}=\sqrt{16448}=\sqrt{64*257}=\sqrt{64}*\sqrt{257}=8\sqrt{257}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{257}}{2*4}=\frac{0-8\sqrt{257}}{8}
=-\frac{8\sqrt{257}}{8}
=-\sqrt{257}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{257}}{2*4}=\frac{0+8\sqrt{257}}{8}
=\frac{8\sqrt{257}}{8}
=\sqrt{257}
$