2(4x+3)(3x+5)=x2+58x+

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Solution for 2(4x+3)(3x+5)=x2+58x+ equation:



2(4x+3)(3x+5)=x2+58x+
We move all terms to the left:
2(4x+3)(3x+5)-(x2+58x+)=0
We add all the numbers together, and all the variables
-(+x^2+58x+)+2(4x+3)(3x+5)=0
We get rid of parentheses
-x^2-58x+2(4x+3)(3x+5)-=0
We multiply parentheses ..
-x^2+2(+12x^2+20x+9x+15)-58x-=0
We add all the numbers together, and all the variables
-1x^2+2(+12x^2+20x+9x+15)-58x=0
We multiply parentheses
-1x^2+24x^2+40x+18x-58x+30=0
We add all the numbers together, and all the variables
23x^2+30=0
a = 23; b = 0; c = +30;
Δ = b2-4ac
Δ = 02-4·23·30
Δ = -2760
Delta is less than zero, so there is no solution for the equation

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