1x^2+17x+42=0

Solution for 1x^2+17x+42=0 equation:

1x^2+17x+42=0

We add all the numbers together, and all the variables

x^2+17x+42=0

a = 1; b = 17; c = +42;
Δ = b2-4ac
Δ = 172-4·1·42
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-11}{2*1}=\frac{-28}{2}
=-14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+11}{2*1}=\frac{-6}{2}
=-3
$