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3x+10=(x+4)(x+4)
We move all terms to the left:
3x+10-((x+4)(x+4))=0
We multiply parentheses ..
-((+x^2+4x+4x+16))+3x+10=0
We calculate terms in parentheses: -((+x^2+4x+4x+16)), so:We add all the numbers together, and all the variables
(+x^2+4x+4x+16)
We get rid of parentheses
x^2+4x+4x+16
We add all the numbers together, and all the variables
x^2+8x+16
Back to the equation:
-(x^2+8x+16)
3x-(x^2+8x+16)+10=0
We get rid of parentheses
-x^2+3x-8x-16+10=0
We add all the numbers together, and all the variables
-1x^2-5x-6=0
a = -1; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·(-1)·(-6)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*-1}=\frac{4}{-2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*-1}=\frac{6}{-2} =-3 $
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