5t^2+6t-43=0

Solution for 5t^2+6t-43=0 equation:

5t^2+6t-43=0

a = 5; b = 6; c = -43;
Δ = b2-4ac
Δ = 62-4·5·(-43)
Δ = 896
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:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-8\sqrt{14}}{2*5}=\frac{-6-8\sqrt{14}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+8\sqrt{14}}{2*5}=\frac{-6+8\sqrt{14}}{10}
$