147t^2-48=0

Solution for 147t^2-48=0 equation:

147t^2-48=0

a = 147; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·147·(-48)
Δ = 28224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{28224}=168$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-168}{2*147}=\frac{-168}{294}
=-4/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+168}{2*147}=\frac{168}{294}
=4/7
$