4.9t^2+26t-15=0

Solution for 4.9t^2+26t-15=0 equation:

4.9t^2+26t-15=0

a = 4.9; b = 26; c = -15;
Δ = b2-4ac
Δ = 262-4·4.9·(-15)
Δ = 970
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{-(26)-\sqrt{970}}{2*4.9}=\frac{-26-\sqrt{970}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+\sqrt{970}}{2*4.9}=\frac{-26+\sqrt{970}}{9.8}
$