3x^2+17x-6.1=0

Solution for 3x^2+17x-6.1=0 equation:

3x^2+17x-6.1=0

a = 3; b = 17; c = -6.1;
Δ = b2-4ac
Δ = 172-4·3·(-6.1)
Δ = 362.2
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{-(17)-\sqrt{362.2}}{2*3}=\frac{-17-\sqrt{362.2}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{362.2}}{2*3}=\frac{-17+\sqrt{362.2}}{6}
$