How do you solve $4x^2+23x-10=10x + x^2$ by using the quadratic formula?

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Answer 1 · 2017-03-07T12:29:38

$x= -5$ or $x=2/3$

$4x^2+23x-10= 10x+x^2$

$3x^2+13x-10=0$

Because the question asks us use the quadratic formula:

$x= (-13+-sqrt(13^2-4xx3xx(-10)))/(2xx3)$

$= (-13+-sqrt(169 + 120))/(2xx3)$

$= (-13+-sqrt(289))/6$

$= (-13+-17)/6$

$= -5$ or $2/3$

But also consider:

$3x^2+13x-10=0$

$(x+5)(3x-2) = 0 -> x= -5$ or $2/3$

Therefore we could have arrived at the result without the quadratic formula in this case.

How do you solve 4x^2+23x-10=10x + x^24x^2+23x-10=10x + x^2 by using the quadratic formula?