2x^2+19x-9.3=0

Solution for 2x^2+19x-9.3=0 equation:

2x^2+19x-9.3=0

a = 2; b = 19; c = -9.3;
Δ = b2-4ac
Δ = 192-4·2·(-9.3)
Δ = 435.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{435.4}}{2*2}=\frac{-19-\sqrt{435.4}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{435.4}}{2*2}=\frac{-19+\sqrt{435.4}}{4}
$