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

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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#