2t^2+14t-20=0

Solution for 2t^2+14t-20=0 equation:

2t^2+14t-20=0

a = 2; b = 14; c = -20;
Δ = b2-4ac
Δ = 142-4·2·(-20)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{89}}{2*2}=\frac{-14-2\sqrt{89}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{89}}{2*2}=\frac{-14+2\sqrt{89}}{4}
$