2t^2+10t=3

Solution for 2t^2+10t=3 equation:

2t^2+10t=3

We move all terms to the left:

2t^2+10t-(3)=0

a = 2; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·2·(-3)
Δ = 124
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:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{31}}{2*2}=\frac{-10-2\sqrt{31}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{31}}{2*2}=\frac{-10+2\sqrt{31}}{4}
$