4t^2+20t+7=0

Solution for 4t^2+20t+7=0 equation:

4t^2+20t+7=0

a = 4; b = 20; c = +7;
Δ = b2-4ac
Δ = 202-4·4·7
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{2}}{2*4}=\frac{-20-12\sqrt{2}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{2}}{2*4}=\frac{-20+12\sqrt{2}}{8}
$