5t^2+35t-35=0

Solution for 5t^2+35t-35=0 equation:

5t^2+35t-35=0

a = 5; b = 35; c = -35;
Δ = b2-4ac
Δ = 352-4·5·(-35)
Δ = 1925
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:

$\sqrt{\Delta}=\sqrt{1925}=\sqrt{25*77}=\sqrt{25}*\sqrt{77}=5\sqrt{77}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5\sqrt{77}}{2*5}=\frac{-35-5\sqrt{77}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5\sqrt{77}}{2*5}=\frac{-35+5\sqrt{77}}{10}
$