3x^2+25x-6.3=0

Solution for 3x^2+25x-6.3=0 equation:

3x^2+25x-6.3=0

a = 3; b = 25; c = -6.3;
Δ = b2-4ac
Δ = 252-4·3·(-6.3)
Δ = 700.6
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{-(25)-\sqrt{700.6}}{2*3}=\frac{-25-\sqrt{700.6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{700.6}}{2*3}=\frac{-25+\sqrt{700.6}}{6}
$