4t^2-112t+490=0

Solution for 4t^2-112t+490=0 equation:

4t^2-112t+490=0

a = 4; b = -112; c = +490;
Δ = b2-4ac
Δ = -1122-4·4·490
Δ = 4704
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:

$\sqrt{\Delta}=\sqrt{4704}=\sqrt{784*6}=\sqrt{784}*\sqrt{6}=28\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-28\sqrt{6}}{2*4}=\frac{112-28\sqrt{6}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+28\sqrt{6}}{2*4}=\frac{112+28\sqrt{6}}{8}
$