10t-16t^2+675=0

Solution for 10t-16t^2+675=0 equation:

10t-16t^2+675=0

a = -16; b = 10; c = +675;
Δ = b2-4ac
Δ = 102-4·(-16)·675
Δ = 43300
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:

$\sqrt{\Delta}=\sqrt{43300}=\sqrt{100*433}=\sqrt{100}*\sqrt{433}=10\sqrt{433}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{433}}{2*-16}=\frac{-10-10\sqrt{433}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{433}}{2*-16}=\frac{-10+10\sqrt{433}}{-32}
$