25t^2+46t-2=0

Solution for 25t^2+46t-2=0 equation:

25t^2+46t-2=0

a = 25; b = 46; c = -2;
Δ = b2-4ac
Δ = 462-4·25·(-2)
Δ = 2316
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:

$\sqrt{\Delta}=\sqrt{2316}=\sqrt{4*579}=\sqrt{4}*\sqrt{579}=2\sqrt{579}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-2\sqrt{579}}{2*25}=\frac{-46-2\sqrt{579}}{50}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+2\sqrt{579}}{2*25}=\frac{-46+2\sqrt{579}}{50}
$