(4x-3)(2/3x+5)=0

Solution for (4x-3)(2/3x+5)=0 equation:

(4x-3)(2/3x+5)=0


Domain of the equation: 3x+5)!=0

x∈R


We multiply parentheses ..

(+8x^2+20x-6x-15)=0

We get rid of parentheses

8x^2+20x-6x-15=0

We add all the numbers together, and all the variables

8x^2+14x-15=0

a = 8; b = 14; c = -15;
Δ = b2-4ac
Δ = 142-4·8·(-15)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-26}{2*8}=\frac{-40}{16}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+26}{2*8}=\frac{12}{16}
=3/4
$