2t^2+2t-87=0

Solution for 2t^2+2t-87=0 equation:

2t^2+2t-87=0

a = 2; b = 2; c = -87;
Δ = b2-4ac
Δ = 22-4·2·(-87)
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10\sqrt{7}}{2*2}=\frac{-2-10\sqrt{7}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10\sqrt{7}}{2*2}=\frac{-2+10\sqrt{7}}{4}
$