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How do you solve #-2x^2-7x=-1.5# using the quadratic formula?

Answer 1

Nathan Axel

#x~~ -3.7,x~~ 0.2#

#"rearrange and equate to zero"#

#rArr-2x^2-7x+1.5=0#

#"with "a=-2,b=-7,c=1.5#

#"solving using the "color(blue)"quadratic formula"#

#"for the quadratic "ax^2+bx+c=0#

#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#

#rArrx=(7+-sqrt(49+12))/(-4)#

#rArrx=(7+sqrt61)/(-4)" or "x=(7-sqrt61)/(-4)#

#rArrx~~ -3.7" to 1 dec. place or "x~~ 0.2" to 1 dec. place"#

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Answer 2

Isaiah Anthony

To solve the quadratic equation -2x^2 - 7x = -1.5 using the quadratic formula, first, identify the coefficients a, b, and c from the equation:

a = -2 b = -7 c = -1.5

Now, plug these values into the quadratic formula:

[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Substitute the values:

[ x = \frac{{-(-7) \pm \sqrt{{(-7)^2 - 4(-2)(-1.5)}}}}{{2(-2)}} ]

[ x = \frac{{7 \pm \sqrt{{49 - 12}}}}{{-4}} ]

[ x = \frac{{7 \pm \sqrt{37}}}{{-4}} ]

So, the solutions for the equation -2x^2 - 7x = -1.5 are:

[ x = \frac{{7 + \sqrt{37}}}{{-4}} ] and [ x = \frac{{7 - \sqrt{37}}}{{-4}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.