t^2+15t=448

Solution for t^2+15t=448 equation:

t^2+15t=448

We move all terms to the left:

t^2+15t-(448)=0

a = 1; b = 15; c = -448;
Δ = b2-4ac
Δ = 152-4·1·(-448)
Δ = 2017
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{2017}}{2*1}=\frac{-15-\sqrt{2017}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{2017}}{2*1}=\frac{-15+\sqrt{2017}}{2}
$