How do you solve $y = 3x^2 + 5x$ using the quadratic formula?

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Answer 1 · 2018-06-17T00:27:33

$x=0,$ $-5/3$

$y=3x^2+5x$ is a quadratic equation in standard form:

$y=ax^2+bx+c$ ,

where:

$a=3$ , $b=5$ , $c=0$

To solve the equation for $x$ , substitute $0$ for $y$ .

$0=3x^2+5x$

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

Plug in the known values and solve.

$x=(-5+-sqrt(5^2-4*3*0))/(2*3)$

$x=(-5+-sqrt25)/6$

$x=(-5+-5)/6$

$x=(-5+5)/6,$ $(-5-5)/6$

$x=0/6, -10/6$

Simplify.

$x=0,$ $-5/3$

How do you solve y = 3x^2 + 5xy = 3x^2 + 5x using the quadratic formula?