5t^2-70t+35=0

Solution for 5t^2-70t+35=0 equation:

5t^2-70t+35=0

a = 5; b = -70; c = +35;
Δ = b2-4ac
Δ = -702-4·5·35
Δ = 4200
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:

$\sqrt{\Delta}=\sqrt{4200}=\sqrt{100*42}=\sqrt{100}*\sqrt{42}=10\sqrt{42}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-10\sqrt{42}}{2*5}=\frac{70-10\sqrt{42}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+10\sqrt{42}}{2*5}=\frac{70+10\sqrt{42}}{10}
$