-2t^2+28t=80

Solution for -2t^2+28t=80 equation:

-2t^2+28t=80

We move all terms to the left:

-2t^2+28t-(80)=0

a = -2; b = 28; c = -80;
Δ = b2-4ac
Δ = 282-4·(-2)·(-80)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12}{2*-2}=\frac{-40}{-4}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12}{2*-2}=\frac{-16}{-4}
=+4
$