How do you solve $-3x^2+26x=16$ ?

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Answer 1 · 2017-02-13T22:58:17

The roots of the quadratic are 0.67 and 8

Start by rearranging the equation into the standard form

$-3x^2 + 26x -16 = 0$

Using the quadratic formula $(-b +- sqrt(b^2 - 4ac))/(2a)$

substitute in the values from the equation
$(-26 +- sqrt(26^2 - 4*-3*-16))/(2*-3)$

Simplifying
$(-26 +- sqrt(676- 192))/-6$

$(-26 +- 22)/-6$

Using the positive discriminant (+22) the root is 0.67
Using the negative discriminant (-22), the root is 8

How do you solve -3x^2+26x=16-3x^2+26x=16?