63t^2-48=0

Solution for 63t^2-48=0 equation:

63t^2-48=0

a = 63; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·63·(-48)
Δ = 12096
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:

$\sqrt{\Delta}=\sqrt{12096}=\sqrt{576*21}=\sqrt{576}*\sqrt{21}=24\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{21}}{2*63}=\frac{0-24\sqrt{21}}{126}
=-\frac{24\sqrt{21}}{126}
=-\frac{4\sqrt{21}}{21}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{21}}{2*63}=\frac{0+24\sqrt{21}}{126}
=\frac{24\sqrt{21}}{126}
=\frac{4\sqrt{21}}{21}
$