4.9t^2+14t=23

Solution for 4.9t^2+14t=23 equation:

4.9t^2+14t=23

We move all terms to the left:

4.9t^2+14t-(23)=0

a = 4.9; b = 14; c = -23;
Δ = b2-4ac
Δ = 142-4·4.9·(-23)
Δ = 646.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-\sqrt{646.8}}{2*4.9}=\frac{-14-\sqrt{646.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+\sqrt{646.8}}{2*4.9}=\frac{-14+\sqrt{646.8}}{9.8}
$