t^2+39=97

Solution for t^2+39=97 equation:

t^2+39=97

We move all terms to the left:

t^2+39-(97)=0

We add all the numbers together, and all the variables

t^2-58=0

a = 1; b = 0; c = -58;
Δ = b2-4ac
Δ = 02-4·1·(-58)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{58}}{2*1}=\frac{0-2\sqrt{58}}{2}
=-\frac{2\sqrt{58}}{2}
=-\sqrt{58}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{58}}{2*1}=\frac{0+2\sqrt{58}}{2}
=\frac{2\sqrt{58}}{2}
=\sqrt{58}
$