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10x(x+3)-(9x-4)=x-5+3
We move all terms to the left:
10x(x+3)-(9x-4)-(x-5+3)=0
We add all the numbers together, and all the variables
10x(x+3)-(9x-4)-(x-2)=0
We multiply parentheses
10x^2+30x-(9x-4)-(x-2)=0
We get rid of parentheses
10x^2+30x-9x-x+4+2=0
We add all the numbers together, and all the variables
10x^2+20x+6=0
a = 10; b = 20; c = +6;
Δ = b2-4ac
Δ = 202-4·10·6
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{10}}{2*10}=\frac{-20-4\sqrt{10}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{10}}{2*10}=\frac{-20+4\sqrt{10}}{20} $
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