2t^2+4t-7=0

Solution for 2t^2+4t-7=0 equation:

2t^2+4t-7=0

a = 2; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·2·(-7)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{2}}{2*2}=\frac{-4-6\sqrt{2}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{2}}{2*2}=\frac{-4+6\sqrt{2}}{4}
$