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How do you solve #2x ^ { 2} = 19x - 3#?

Question

How do you solve #2x ^ { 2} = 19x - 3#?
How do you solve #2x ^ { 2} = 19x - 3#?

Answers

well, you just...

Explanation:

...move the terms on the right to the left, and that gives you a nice quadratic equation to solve.

Subtract 19x from both sides:

#2x^2-19x = -3#

...then add 3 to both sides:

#2x^2-19x+3 = 0#

...starting to look familiar?

If #ax^2 + bx + c = 0#, then

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

...so here, a = 2, b = -19, c = 3

#x = (19 +-sqrt((-19^2)-(4 * 2 * 3)))/(2*2)#

#x = (19 +- sqrt(337))/4#

x = 9.339 (rounding) or x = 0.161 (rounding)

GOOD LUCK!


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