What are the solutions of 3x^2 +6x +15 =0

Step-by-step explanation:

Use the quadratic formula

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x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c{2{\color{#c92786}{a}}}

x=2a−b±b2−4ac

Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.

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3x^{2}+6x+15=0

3x2+6x+15=0

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a={\color{#c92786}{3}}

a=3

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b={\color{#e8710a}{6}}

b=6

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c={\color{#129eaf}{15}}

c=15

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x=\frac{-{\color{#e8710a}{6}} \pm \sqrt{{\color{#e8710a}{6}}^{2}-4 \cdot {\color{#c92786}{3}} \cdot {\color{#129eaf}{15{2 \cdot {\color{#c92786}{3}}}

x=2⋅3−6±62−4⋅3⋅15

2

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?