5t^2+4t-7=0

Solution for 5t^2+4t-7=0 equation:

5t^2+4t-7=0

a = 5; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·5·(-7)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{39}}{2*5}=\frac{-4-2\sqrt{39}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{39}}{2*5}=\frac{-4+2\sqrt{39}}{10}
$