-3e2+10e+21=0

Solution for -3e2+10e+21=0 equation:

-3e^2+10e+21=0

a = -3; b = 10; c = +21;
Δ = b2-4ac
Δ = 102-4·(-3)·21
Δ = 352
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{22}}{2*-3}=\frac{-10-4\sqrt{22}}{-6}
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{22}}{2*-3}=\frac{-10+4\sqrt{22}}{-6}
$