t^2+8t=48

Solution for t^2+8t=48 equation:

t^2+8t=48

We move all terms to the left:

t^2+8t-(48)=0

a = 1; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·1·(-48)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*1}=\frac{-24}{2}
=-12
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*1}=\frac{8}{2}
=4
$