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

Question

1159 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-02T08:22:27

$x=-3/4+1/4 sqrt(217)$ and $x=-3/4-1/4 sqrt(217)$

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

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

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 $6$ from both sides:

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

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

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

$x=(-3+-sqrt(3^2-4(2)(-26)))/(2(2))$

$x=(-3+-sqrt(217))/(4)$

This gives two solutions

$x=-3/4+1/4 sqrt(217)$ and $x=-3/4-1/4 sqrt(217)$

In decimal form, we have

$x~~2.9327$ and $x~~-4.4327$

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