2t2+8t+7=0

Solution for 2t2+8t+7=0 equation:

2t^2+8t+7=0

a = 2; b = 8; c = +7;
Δ = b2-4ac
Δ = 82-4·2·7
Δ = 8
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:

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