13x^2+25x+12=0

Solution for 13x^2+25x+12=0 equation:

13x^2+25x+12=0

a = 13; b = 25; c = +12;
Δ = b2-4ac
Δ = 252-4·13·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-1}{2*13}=\frac{-26}{26}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+1}{2*13}=\frac{-24}{26}
=-12/13
$