10t^2+23t+6=0

Solution for 10t^2+23t+6=0 equation:

10t^2+23t+6=0

a = 10; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·10·6
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*10}=\frac{-40}{20}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*10}=\frac{-6}{20}
=-3/10
$