How do you solve $y = 3 x^{2} - 2 x - 5$ $y = 3x^2 − 2x − 5$ ?

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Answer 1 · 2017-09-07T17:30:26

$3 x^{2} - 2 x - 5 = 0$ $3x^2-2x-5=0$

$3 x^{2} + 3 x - 5 x - 5 = 0$ $3x^2+3x-5x-5=0$

$3 x \cdot (x + 1) - 5 \cdot (x + 1) = 0$ $3x*(x+1)-5*(x+1)=0$

$(x + 1) \cdot (3 x - 5) = 0$ $(x+1)*(3x-5)=0$

Thus, $x 1 = - 1$ $x1=-1$ and $x 2 = \frac{5}{3}$ $x2=5/3$ .

I used distribution and collection rules of polynomials.

How do you solve y=3x2−2x−5y = 3x^2 − 2x − 5 ?