How do you solve using the quadratic formula $17 x^{2} = 12 x$ $17x^2 = 12x$ ?

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How do you solve using the quadratic formula $17 x^{2} = 12 x$ $17x^2 = 12x$ ?

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Answer 1 · 2015-04-29T13:20:19

To use the quadratic formula (which, by the way is not the easiest way to do this) first convert into the standard form
$a x^{2} + b x + c = 0$ $ax^2+bx+c =0$

$17 x^{2} = 12 x$ $17x^2=12x$
$17 x^{2} - 12 x + 0 = 0$ $17x^2-12x+0=0$

The quadratic formula for roots is
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{12 \pm \sqrt{{(- 12)}^{2} - 4 (17) (0)}}{2 (17)}$ $x=(12+-sqrt((-12)^2- 4(17)(0)))/(2(17))$

$x = \frac{12}{17}$ $x= 12/17$
or
$x = 0$ $x=0$

Answer 2 · 2015-04-29T13:24:06

Write it as:
$17 x^{2} - 12 x = 0$ $17x^2-12x=0$
this is now in the form: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
With:
$a = 17$ $a=17$
$b = - 12$ $b=-12$
$c = 0$ $c=0$
That can be used in the quadratic formula to get your two Real solutions (if possible) for $x$ $x$ as:
$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$
Try by yourself substituting values (you should get two real values).

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How do you solve using the quadratic formula $17 x^{2} = 12 x$ $17x^2 = 12x$ ?

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