How do you solve $6-7x=3x²$ ?

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Answer 1 · 2017-06-22T08:04:29

$x=2/3, x=-3$

Put all terms onto one side of the equation so that everything is equal to $0$ .
$6-7x=3x^2$
$6=3x^2+7x$
$0=3x^2+7x-6$

Factor.
$3x^2+7x-6=0$
$(3x-2)(x+3)=0$

Set each factor equal to $0$ . Make sure to do this for both.
$3x-2=0$
$3x=2$
$x=2/3$

$x+3=0$
$x=-3$

Answers: $x=2/3, x=-3$

Answer 2 · 2017-06-22T14:44:19

$2/3 and - 3$

$y = 3x^2 + 7x - 6 = 0$

Use the new transforming method (Socratic Search)
Transformed equation:

$y' = x^2 + 7x - 18 = 0" "rarr (a xx c = 3(-6) = -18)$

Proceed. Find 2 real roots of y', then, divide them by $a = 3.$

Find $2$ numbers knowing the sum $(-b = -7))$
and the product $(a xx c = -18)$ .

They are: $2, and (-9).$

Consequently, the $2$ real roots of $y$ are:

$x_1 = 2/a = 2/3$ , and $x_2 = -9/3 = -3$

Note. This new method avoids the lengthy factoring by grouping and solving the 2 binomials.

How do you solve 6-7x=3x²6-7x=3x²?