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(3+5x)(2-3x)=12-5x^2
We move all terms to the left:
(3+5x)(2-3x)-(12-5x^2)=0
We add all the numbers together, and all the variables
-(12-5x^2)+(5x+3)(-3x+2)=0
We get rid of parentheses
5x^2+(5x+3)(-3x+2)-12=0
We multiply parentheses ..
5x^2+(-15x^2+10x-9x+6)-12=0
We get rid of parentheses
5x^2-15x^2+10x-9x+6-12=0
We add all the numbers together, and all the variables
-10x^2+x-6=0
a = -10; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·(-10)·(-6)
Δ = -239
Delta is less than zero, so there is no solution for the equation
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