3x^2+7x-440=0

Solution for 3x^2+7x-440=0 equation:

3x^2+7x-440=0

a = 3; b = 7; c = -440;
Δ = b2-4ac
Δ = 72-4·3·(-440)
Δ = 5329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5329}=73$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-73}{2*3}=\frac{-80}{6}
=-13+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+73}{2*3}=\frac{66}{6}
=11
$