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(3x+1)(3x+1)+(x+2)(x+2)=65
We move all terms to the left:
(3x+1)(3x+1)+(x+2)(x+2)-(65)=0
We multiply parentheses ..
(+9x^2+3x+3x+1)+(x+2)(x+2)-65=0
We get rid of parentheses
9x^2+3x+3x+(x+2)(x+2)+1-65=0
We multiply parentheses ..
9x^2+(+x^2+2x+2x+4)+3x+3x+1-65=0
We add all the numbers together, and all the variables
9x^2+(+x^2+2x+2x+4)+6x-64=0
We get rid of parentheses
9x^2+x^2+2x+2x+6x+4-64=0
We add all the numbers together, and all the variables
10x^2+10x-60=0
a = 10; b = 10; c = -60;
Δ = b2-4ac
Δ = 102-4·10·(-60)
Δ = 2500
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{2500}=50$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-50}{2*10}=\frac{-60}{20} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+50}{2*10}=\frac{40}{20} =2 $
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