4*x^2=1400

Solution for 4*x^2=1400 equation:

4x^2=1400

We move all terms to the left:

4x^2-(1400)=0

a = 4; b = 0; c = -1400;
Δ = b2-4ac
Δ = 02-4·4·(-1400)
Δ = 22400
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:

$\sqrt{\Delta}=\sqrt{22400}=\sqrt{1600*14}=\sqrt{1600}*\sqrt{14}=40\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{14}}{2*4}=\frac{0-40\sqrt{14}}{8}
=-\frac{40\sqrt{14}}{8}
=-5\sqrt{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{14}}{2*4}=\frac{0+40\sqrt{14}}{8}
=\frac{40\sqrt{14}}{8}
=5\sqrt{14}
$