3t^2-19-40=0

Solution for 3t^2-19-40=0 equation:

3t^2-19-40=0

We add all the numbers together, and all the variables

3t^2-59=0

a = 3; b = 0; c = -59;
Δ = b2-4ac
Δ = 02-4·3·(-59)
Δ = 708
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:

$\sqrt{\Delta}=\sqrt{708}=\sqrt{4*177}=\sqrt{4}*\sqrt{177}=2\sqrt{177}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{177}}{2*3}=\frac{0-2\sqrt{177}}{6}
=-\frac{2\sqrt{177}}{6}
=-\frac{\sqrt{177}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{177}}{2*3}=\frac{0+2\sqrt{177}}{6}
=\frac{2\sqrt{177}}{6}
=\frac{\sqrt{177}}{3}
$