3t^2+47=53

Solution for 3t^2+47=53 equation:

3t^2+47=53

We move all terms to the left:

3t^2+47-(53)=0

We add all the numbers together, and all the variables

3t^2-6=0

a = 3; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·3·(-6)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*3}=\frac{0-6\sqrt{2}}{6}
=-\frac{6\sqrt{2}}{6}
=-\sqrt{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*3}=\frac{0+6\sqrt{2}}{6}
=\frac{6\sqrt{2}}{6}
=\sqrt{2}
$