t^2=180/16

Solution for t^2=180/16 equation:

t^2=180/16

We move all terms to the left:

t^2-(180/16)=0

We add all the numbers together, and all the variables

t^2-(+180/16)=0

We get rid of parentheses

t^2-180/16=0

We multiply all the terms by the denominator


t^2*16-180=0

Wy multiply elements

16t^2-180=0

a = 16; b = 0; c = -180;
Δ = b2-4ac
Δ = 02-4·16·(-180)
Δ = 11520
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:

$\sqrt{\Delta}=\sqrt{11520}=\sqrt{2304*5}=\sqrt{2304}*\sqrt{5}=48\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{5}}{2*16}=\frac{0-48\sqrt{5}}{32}
=-\frac{48\sqrt{5}}{32}
=-\frac{3\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{5}}{2*16}=\frac{0+48\sqrt{5}}{32}
=\frac{48\sqrt{5}}{32}
=\frac{3\sqrt{5}}{2}
$