-5t2+10t+3=1

Solution for -5t2+10t+3=1 equation:

-5t^2+10t+3=1

We move all terms to the left:

-5t^2+10t+3-(1)=0

We add all the numbers together, and all the variables

-5t^2+10t+2=0

a = -5; b = 10; c = +2;
Δ = b2-4ac
Δ = 102-4·(-5)·2
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{35}}{2*-5}=\frac{-10-2\sqrt{35}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{35}}{2*-5}=\frac{-10+2\sqrt{35}}{-10}
$