4t^2+9t^2=9

Solution for 4t^2+9t^2=9 equation:

4t^2+9t^2=9

We move all terms to the left:

4t^2+9t^2-(9)=0

We add all the numbers together, and all the variables

13t^2-9=0

a = 13; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·13·(-9)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{13}}{2*13}=\frac{0-6\sqrt{13}}{26}
=-\frac{6\sqrt{13}}{26}
=-\frac{3\sqrt{13}}{13}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{13}}{2*13}=\frac{0+6\sqrt{13}}{26}
=\frac{6\sqrt{13}}{26}
=\frac{3\sqrt{13}}{13}
$