16t^2+30t=270

Solution for 16t^2+30t=270 equation:

16t^2+30t=270

We move all terms to the left:

16t^2+30t-(270)=0

a = 16; b = 30; c = -270;
Δ = b2-4ac
Δ = 302-4·16·(-270)
Δ = 18180
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:

$\sqrt{\Delta}=\sqrt{18180}=\sqrt{36*505}=\sqrt{36}*\sqrt{505}=6\sqrt{505}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6\sqrt{505}}{2*16}=\frac{-30-6\sqrt{505}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6\sqrt{505}}{2*16}=\frac{-30+6\sqrt{505}}{32}
$