4t^2+4t=48

Solution for 4t^2+4t=48 equation:

4t^2+4t=48

We move all terms to the left:

4t^2+4t-(48)=0

a = 4; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·4·(-48)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*4}=\frac{-32}{8}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*4}=\frac{24}{8}
=3
$