10x^2+13x-42=0

Solution for 10x^2+13x-42=0 equation:

10x^2+13x-42=0

a = 10; b = 13; c = -42;
Δ = b2-4ac
Δ = 132-4·10·(-42)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-43}{2*10}=\frac{-56}{20}
=-2+4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+43}{2*10}=\frac{30}{20}
=1+1/2
$