150=400(t)-4.9(t)^2

Solution for 150=400(t)-4.9(t)^2 equation:

150=400(t)-4.9(t)^2

We move all terms to the left:

150-(400(t)-4.9(t)^2)=0

We get rid of parentheses

4.9t^2-400t+150=0

a = 4.9; b = -400; c = +150;
Δ = b2-4ac
Δ = -4002-4·4.9·150
Δ = 157060
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:

$\sqrt{\Delta}=\sqrt{157060}=\sqrt{4*39265}=\sqrt{4}*\sqrt{39265}=2\sqrt{39265}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-400)-2\sqrt{39265}}{2*4.9}=\frac{400-2\sqrt{39265}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-400)+2\sqrt{39265}}{2*4.9}=\frac{400+2\sqrt{39265}}{9.8}
$