4t^2=24+28

Solution for 4t^2=24+28 equation:

4t^2=24+28

We move all terms to the left:

4t^2-(24+28)=0

We add all the numbers together, and all the variables

4t^2-52=0

a = 4; b = 0; c = -52;
Δ = b2-4ac
Δ = 02-4·4·(-52)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{13}}{2*4}=\frac{0-8\sqrt{13}}{8}
=-\frac{8\sqrt{13}}{8}
=-\sqrt{13}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{13}}{2*4}=\frac{0+8\sqrt{13}}{8}
=\frac{8\sqrt{13}}{8}
=\sqrt{13}
$