4t^2+15t=14

Solution for 4t^2+15t=14 equation:

4t^2+15t=14

We move all terms to the left:

4t^2+15t-(14)=0

a = 4; b = 15; c = -14;
Δ = b2-4ac
Δ = 152-4·4·(-14)
Δ = 449
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{-(15)-\sqrt{449}}{2*4}=\frac{-15-\sqrt{449}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{449}}{2*4}=\frac{-15+\sqrt{449}}{8}
$