How do you solve the equation $3 x^{2} + 6 x - 1 = 8$ $3x^2+6x-1=8$ ?

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How do you solve the equation $3 x^{2} + 6 x - 1 = 8$ $3x^2+6x-1=8$ ?

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Answer 1 · 2015-02-04T22:11:37

You want an equation like $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , so we subtract 8 from both sides. We now get $3 x^{2} + 6 x - 9 = 0$ $3x^2+6x-9 = 0$ , so we find $a = 3, b = 6$ $a=3, b=6$ and $c = - 9$ $c=-9$ .
We use the quadratic formula to solve this equation:
$D = b^{2} - 4 a c = 6^{2} - 4 \cdot 3 \cdot - 9 = 36 + 108 = 144 .$ $D=b^2-4ac = 6^2-4*3*-9 = 36+108=144.$
We know that $x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{- 6 \pm \sqrt{144}}{2 \cdot 3} = \frac{- 6 \pm 12}{6} = - 1 \pm 2$ $x=(-b+-sqrt(D))/(2a) = (-6+-sqrt(144))/(2*3) = (-6+-12)/6= -1+-2$ , so $x = 1 \lor x = - 3$ $x=1 vee x=-3$ .

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How do you solve the equation $3 x^{2} + 6 x - 1 = 8$ $3x^2+6x-1=8$ ?

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