Question

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Answer 1

x^2-5x-5 = 0

The equation is in the form:

ax^2+bx+c

Where:

a= 1

b= -5

c= -5

Apply the quadratic formula:

$\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2\cdot a}$

Replacing:

$\frac{-(-5)\pm\sqrt[]{(-5)^2-4\cdot1\cdot-5}}{2\cdot1}$$\frac{5\pm\sqrt[]{25+20}}{2}$$\frac{5\pm\sqrt[]{45}}{2}$$\frac{5\pm3\sqrt[]{5}}{2}$

Positive:

(5+√45)/2 = (5+3√5)/2 = 5/2+3/2√5

Negative

(5-√45) /2 = (5-3√5)/2=5/2-3/2√5