(4x-5)(3x+14)=0

Solution for (4x-5)(3x+14)=0 equation:

(4x-5)(3x+14)=0

We multiply parentheses ..

(+12x^2+56x-15x-70)=0

We get rid of parentheses

12x^2+56x-15x-70=0

We add all the numbers together, and all the variables

12x^2+41x-70=0

a = 12; b = 41; c = -70;
Δ = b2-4ac
Δ = 412-4·12·(-70)
Δ = 5041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5041}=71$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-71}{2*12}=\frac{-112}{24}
=-4+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+71}{2*12}=\frac{30}{24}
=1+1/4
$