3w^2+10w-13=0

Solution for 3w^2+10w-13=0 equation:

3w^2+10w-13=0

a = 3; b = 10; c = -13;
Δ = b2-4ac
Δ = 102-4·3·(-13)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16}{2*3}=\frac{-26}{6}
=-4+1/3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16}{2*3}=\frac{6}{6}
=1
$