2(4x+3)(3x+5)=12x^2+58x+30

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Solution for 2(4x+3)(3x+5)=12x^2+58x+30 equation: 



2(4x+3)(3x+5)=12x^2+58x+30

We move all terms to the left:

2(4x+3)(3x+5)-(12x^2+58x+30)=0

We get rid of parentheses

-12x^2+2(4x+3)(3x+5)-58x-30=0

We multiply parentheses ..

-12x^2+2(+12x^2+20x+9x+15)-58x-30=0

We multiply parentheses

-12x^2+24x^2+40x+18x-58x+30-30=0

We add all the numbers together, and all the variables

12x^2=0

a = 12; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·12·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{24}=0$

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