3t^2-46t+15=0

Solution for 3t^2-46t+15=0 equation:

3t^2-46t+15=0

a = 3; b = -46; c = +15;
Δ = b2-4ac
Δ = -462-4·3·15
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-44}{2*3}=\frac{2}{6}
=1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+44}{2*3}=\frac{90}{6}
=15
$