4t^2=24t+28

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

4t^2=24t+28

We move all terms to the left:

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

We get rid of parentheses

4t^2-24t-28=0

a = 4; b = -24; c = -28;
Δ = b2-4ac
Δ = -242-4·4·(-28)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-32}{2*4}=\frac{-8}{8}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+32}{2*4}=\frac{56}{8}
=7
$