2t^2+4t-78=0

Solution for 2t^2+4t-78=0 equation:

2t^2+4t-78=0

a = 2; b = 4; c = -78;
Δ = b2-4ac
Δ = 42-4·2·(-78)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{10}}{2*2}=\frac{-4-8\sqrt{10}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{10}}{2*2}=\frac{-4+8\sqrt{10}}{4}
$