16t^2=30+290t

Solution for 16t^2=30+290t equation:

16t^2=30+290t

We move all terms to the left:

16t^2-(30+290t)=0

We add all the numbers together, and all the variables

16t^2-(290t+30)=0

We get rid of parentheses

16t^2-290t-30=0

a = 16; b = -290; c = -30;
Δ = b2-4ac
Δ = -2902-4·16·(-30)
Δ = 86020
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:

$\sqrt{\Delta}=\sqrt{86020}=\sqrt{4*21505}=\sqrt{4}*\sqrt{21505}=2\sqrt{21505}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-290)-2\sqrt{21505}}{2*16}=\frac{290-2\sqrt{21505}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-290)+2\sqrt{21505}}{2*16}=\frac{290+2\sqrt{21505}}{32}
$