t^2+4t-280=0

Solution for t^2+4t-280=0 equation:

t^2+4t-280=0

a = 1; b = 4; c = -280;
Δ = b2-4ac
Δ = 42-4·1·(-280)
Δ = 1136
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:

$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{71}}{2*1}=\frac{-4-4\sqrt{71}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{71}}{2*1}=\frac{-4+4\sqrt{71}}{2}
$