1.68^2=t^2

Solution for 1.68^2=t^2 equation:

1.68^2=t^2

We move all terms to the left:

1.68^2-(t^2)=0

We add all the numbers together, and all the variables

-1t^2+2.8224=0

a = -1; b = 0; c = +2.8224;
Δ = b2-4ac
Δ = 02-4·(-1)·2.8224
Δ = 11.2896
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{11.2896}}{2*-1}=\frac{0-\sqrt{11.2896}}{-2}
=-\frac{\sqrt{}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{11.2896}}{2*-1}=\frac{0+\sqrt{11.2896}}{-2}
=\frac{\sqrt{}}{-2}
$