36t^2+29t^2-7=0

Solution for 36t^2+29t^2-7=0 equation:

36t^2+29t^2-7=0

We add all the numbers together, and all the variables

65t^2-7=0

a = 65; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·65·(-7)
Δ = 1820
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:

$\sqrt{\Delta}=\sqrt{1820}=\sqrt{4*455}=\sqrt{4}*\sqrt{455}=2\sqrt{455}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{455}}{2*65}=\frac{0-2\sqrt{455}}{130}
=-\frac{2\sqrt{455}}{130}
=-\frac{\sqrt{455}}{65}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{455}}{2*65}=\frac{0+2\sqrt{455}}{130}
=\frac{2\sqrt{455}}{130}
=\frac{\sqrt{455}}{65}
$