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(2x+1)(3x+-5)=
We move all terms to the left:
(2x+1)(3x+-5)-()=0
We add all the numbers together, and all the variables
(2x+1)(3x-5)-()=0
We add all the numbers together, and all the variables
(2x+1)(3x-5)=0
We multiply parentheses ..
(+6x^2-10x+3x-5)=0
We get rid of parentheses
6x^2-10x+3x-5=0
We add all the numbers together, and all the variables
6x^2-7x-5=0
a = 6; b = -7; c = -5;
Δ = b2-4ac
Δ = -72-4·6·(-5)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*6}=\frac{-6}{12} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*6}=\frac{20}{12} =1+2/3 $
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