15x^2+16x+1=0

Solution for 15x^2+16x+1=0 equation:

15x^2+16x+1=0

a = 15; b = 16; c = +1;
Δ = b2-4ac
Δ = 162-4·15·1
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*15}=\frac{-30}{30}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*15}=\frac{-2}{30}
=-1/15
$