Math, asked by viji97b, 15 days ago

12y^2-40y+12=0 solve this​

Answers

Answered by McLarenArnab
0

Answer:

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Answered by sitarasahu42
0

Step-by-step explanation:

How to solve your problem

Topics: Algebra, Quadratic Equation

122−40+12=0

12y^{2}-40y+12=012y2−40y+12=0

Quadratic formula

Factor

1

Common factor

122−40+12=0

12y^{2}-40y+12=012y2−40y+12=0

4(32−10+3)=0

4(3y^{2}-10y+3)=04(3y2−10y+3)=0

2

Divide both sides of the equation by the same term

4(32−10+3)=0

4(3y^{2}-10y+3)=04(3y2−10y+3)=0

32−10+3=0

3y^{2}-10y+3=03y2−10y+3=0

3

Use the quadratic formula

=−±2−4√2

y=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}y=2a−b±b2−4ac

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

32−10+3=0

3y^{2}-10y+3=03y2−10y+3=0

=3

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

=−10

b={\color{#e8710a}{-10}}b=−10

=3

c={\color{#129eaf}{3}}c=3

=−(−10)±(−10)2−4⋅3⋅3√2⋅3

y=\frac{-({\color{#e8710a}{-10}}) \pm \sqrt{({\color{#e8710a}{-10}})^{2}-4 \cdot {\color{#c92786}{3}} \cdot {\color{#129eaf}{3}}}}{2 \cdot {\color{#c92786}{3}}}y=2⋅3−(−10)±(−10)2−4⋅3⋅3

4

Simplify

Evaluate the exponent

Multiply the numbers

Subtract the numbers

Evaluate the square root

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