3.33x^2+3.5x-18=0

Solution for 3.33x^2+3.5x-18=0 equation:

3.33x^2+3.5x-18=0

a = 3.33; b = 3.5; c = -18;
Δ = b2-4ac
Δ = 3.52-4·3.33·(-18)
Δ = 252.01
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{-(3.5)-\sqrt{252.01}}{2*3.33}=\frac{-3.5-\sqrt{252.01}}{6.66}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{252.01}}{2*3.33}=\frac{-3.5+\sqrt{252.01}}{6.66}
$