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2 years ago

What are the solutions to the quadratic equation below? 4x^2 + 28x + 49 = 5

Mathematics

2 answers:

Sveta_85 [38]2 years ago

6 0

Answer:

$x=\frac{-7±\sqrt{5 } }{2}$

Step-by-step explanation:

4x² + 28x + 49 = 5

4x² + 28x + 49 - 5 = 0

4x² + 28x + 44 = 0

4(x² + 7x + 11) = 0

4(x² + 7x + 11) ÷ 4 = 0 ÷ 4

x² + 7x + 11 = 0

$x=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

Ignore the A before the ±. It wouldn't let me type it correctly.

a = 1

b = 7

c = 11

$x=\frac{-7±\sqrt{7^{2}-4((1)(11)) } }{2(1)}$

$x=\frac{-7±\sqrt{49-4((1)(11)) } }{2(1)}$

$x=\frac{-7±\sqrt{49-44 } }{2(1)}$

$x=\frac{-7±\sqrt{5 } }{2(1)}$

$x=\frac{-7±\sqrt{5 } }{2}$

bezimeni [28]2 years ago

4 0

Answer:

Step-by-step explanation:

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Tresset [83]

Answer:

$y - 9 = -\frac{4}{3}(x - 3)$

Step-by-step explanation:

Given the points (3, 9) and (9, 1), we must first solve for the slope of the line before proceeding with writing the point-slope form.

In order to solve for the slope (m ), use the following formula:

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