5t^2=127

Solution for 5t^2=127 equation:

5t^2=127

We move all terms to the left:

5t^2-(127)=0

a = 5; b = 0; c = -127;
Δ = b2-4ac
Δ = 02-4·5·(-127)
Δ = 2540
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2540}=\sqrt{4*635}=\sqrt{4}*\sqrt{635}=2\sqrt{635}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{635}}{2*5}=\frac{0-2\sqrt{635}}{10}
=-\frac{2\sqrt{635}}{10}
=-\frac{\sqrt{635}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{635}}{2*5}=\frac{0+2\sqrt{635}}{10}
=\frac{2\sqrt{635}}{10}
=\frac{\sqrt{635}}{5}
$