5t2+14t-10=0

Solution for 5t2+14t-10=0 equation:

5t^2+14t-10=0

a = 5; b = 14; c = -10;
Δ = b2-4ac
Δ = 142-4·5·(-10)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6\sqrt{11}}{2*5}=\frac{-14-6\sqrt{11}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6\sqrt{11}}{2*5}=\frac{-14+6\sqrt{11}}{10}
$