3t^2+46t-28=0

Solution for 3t^2+46t-28=0 equation:

3t^2+46t-28=0

a = 3; b = 46; c = -28;
Δ = b2-4ac
Δ = 462-4·3·(-28)
Δ = 2452
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:

$\sqrt{\Delta}=\sqrt{2452}=\sqrt{4*613}=\sqrt{4}*\sqrt{613}=2\sqrt{613}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-2\sqrt{613}}{2*3}=\frac{-46-2\sqrt{613}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+2\sqrt{613}}{2*3}=\frac{-46+2\sqrt{613}}{6}
$