148t+16t^2=0

Solution for 148t+16t^2=0 equation:

148t+16t^2=0

a = 16; b = 148; c = 0;
Δ = b2-4ac
Δ = 1482-4·16·0
Δ = 21904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21904}=148$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(148)-148}{2*16}=\frac{-296}{32}
=-9+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(148)+148}{2*16}=\frac{0}{32}
=0
$