-1.2t^2+3.5t+7=0

Solution for -1.2t^2+3.5t+7=0 equation:

-1.2t^2+3.5t+7=0

a = -1.2; b = 3.5; c = +7;
Δ = b2-4ac
Δ = 3.52-4·(-1.2)·7
Δ = 45.85
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{-(3.5)-\sqrt{45.85}}{2*-1.2}=\frac{-3.5-\sqrt{45.85}}{-2.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{45.85}}{2*-1.2}=\frac{-3.5+\sqrt{45.85}}{-2.4}
$