3x^2+10x^2-9x-4=0

Solution for 3x^2+10x^2-9x-4=0 equation:

3x^2+10x^2-9x-4=0

We add all the numbers together, and all the variables

13x^2-9x-4=0

a = 13; b = -9; c = -4;
Δ = b2-4ac
Δ = -92-4·13·(-4)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-17}{2*13}=\frac{-8}{26}
=-4/13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+17}{2*13}=\frac{26}{26}
=1
$