0=5t^2+35t+40

Solution for 0=5t^2+35t+40 equation:

0=5t^2+35t+40

We move all terms to the left:

0-(5t^2+35t+40)=0

We add all the numbers together, and all the variables

-(5t^2+35t+40)=0

We get rid of parentheses

-5t^2-35t-40=0

a = -5; b = -35; c = -40;
Δ = b2-4ac
Δ = -352-4·(-5)·(-40)
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5\sqrt{17}}{2*-5}=\frac{35-5\sqrt{17}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5\sqrt{17}}{2*-5}=\frac{35+5\sqrt{17}}{-10}
$