45=-16t^2+70t

Solution for 45=-16t^2+70t equation:

45=-16t^2+70t

We move all terms to the left:

45-(-16t^2+70t)=0

We get rid of parentheses

16t^2-70t+45=0

a = 16; b = -70; c = +45;
Δ = b2-4ac
Δ = -702-4·16·45
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-2\sqrt{505}}{2*16}=\frac{70-2\sqrt{505}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+2\sqrt{505}}{2*16}=\frac{70+2\sqrt{505}}{32}
$