(100)=0.035t^2+4t

Solution for (100)=0.035t^2+4t equation:

(100)=0.035t^2+4t

We move all terms to the left:

(100)-(0.035t^2+4t)=0

We get rid of parentheses

-0.035t^2-4t+100=0

a = -0.035; b = -4; c = +100;
Δ = b2-4ac
Δ = -42-4·(-0.035)·100
Δ = 30
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-\sqrt{30}}{2*-0.035}=\frac{4-\sqrt{30}}{-0.07}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{30}}{2*-0.035}=\frac{4+\sqrt{30}}{-0.07}
$