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7x+3(x2-5)=x-3
We move all terms to the left:
7x+3(x2-5)-(x-3)=0
We add all the numbers together, and all the variables
3(+x^2-5)+7x-(x-3)=0
We multiply parentheses
3x^2+7x-(x-3)-15=0
We get rid of parentheses
3x^2+7x-x+3-15=0
We add all the numbers together, and all the variables
3x^2+6x-12=0
a = 3; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·3·(-12)
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{5}}{2*3}=\frac{-6-6\sqrt{5}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{5}}{2*3}=\frac{-6+6\sqrt{5}}{6} $
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