How do you solve $3 x^{2} + 4 x = 2$ $3x^2 + 4x = 2$ using the quadratic formula?

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How do you solve $3 x^{2} + 4 x = 2$ $3x^2 + 4x = 2$ using the quadratic formula?

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Answer 1 · 2018-08-03T07:06:54

$x = \frac{- 2 + \sqrt{10}}{3}$ $x = (-2 + sqrt10)/3$ , $x = \frac{- 2 - \sqrt{10}}{3}$ $x = (-2 - sqrt10)/3$

Explanation:

$3 x^{2} + 4 x = 2$ $3x^2 + 4x = 2$

First, make one side equal to zero by subtracting $2$ $color(blue)2$ from both sides:
$3 x^{2} + 4 x - 2 = 0$ $3x^2 + 4x - 2 = 0$

The quadratic formula is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$ , where from our equation we know that $a = 3, b = 4, and c = - 2$ $a = 3, b = 4, and c = -2$ .

Plug them into the formula:
$x = \frac{- 4 \pm \sqrt{4^{2} - 4 (3) (- 2)}}{2 (3)}$ $x = (-4 +- sqrt(4^2 - 4(3)(-2)))/(2(3))$

$x = \frac{- 4 \pm \sqrt{16 + 24}}{6}$ $x = (-4 +- sqrt(16 + 24))/6$

$x = \frac{- 4 \pm \sqrt{40}}{6}$ $x = (-4 +- sqrt40)/6$

$x = \frac{- 4 \pm 2 \sqrt{10}}{6}$ $x = (-4 +- 2sqrt10)/6$

$x = \frac{- 2 \pm \sqrt{10}}{3}$ $x = (-2 +- sqrt10)/3$

Hope this helps!

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How do you solve $3 x^{2} + 4 x = 2$ $3x^2 + 4x = 2$ using the quadratic formula?

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