How do you solve $2 x^{2} + 3 x - 20 = 6$ $2x ^ { 2} + 3x - 20= 6$ ?

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Answer 1 · 2017-06-02T08:22:27

$x = - \frac{3}{4} + \frac{1}{4} \sqrt{217}$ $x=-3/4+1/4 sqrt(217)$ and $x = - \frac{3}{4} - \frac{1}{4} \sqrt{217}$ $x=-3/4-1/4 sqrt(217)$

This can be confusing because as the expression is currently expressed, it can be factored on the left hand side.

$(x + 4) (2 x - 5) = 6$ $(x+4)(2x-5)=6$

However, this is not helpful because we need to know when all the multiplied factors on the left equal zero on the right.

So, to complete this problem, we must subtract $6$ $6$ from both sides:

$2 x^{2} + 3 x - 20 - 6 = 6 - 6$ $2x^2+3x-20color(red)(-6)=6color(red)(-6)$

$2 x^{2} + 3 x - 26 = 0$ $2x^2+3x-26=0$

This does not factor, so we have to use the quadratic equation:

$x = \frac{- 3 \pm \sqrt{3^{2} - 4 (2) (- 26)}}{2 (2)}$ $x=(-3+-sqrt(3^2-4(2)(-26)))/(2(2))$

$x = \frac{- 3 \pm \sqrt{217}}{4}$ $x=(-3+-sqrt(217))/(4)$

This gives two solutions

$x = - \frac{3}{4} + \frac{1}{4} \sqrt{217}$ $x=-3/4+1/4 sqrt(217)$ and $x = - \frac{3}{4} - \frac{1}{4} \sqrt{217}$ $x=-3/4-1/4 sqrt(217)$

In decimal form, we have

$x \approx 2.9327$ $x~~2.9327$ and $x \approx - 4.4327$ $x~~-4.4327$

How do you solve 2x2+3x−20=62x ^ { 2} + 3x - 20= 6?