What are the solutions to the equation 4p^2-4p-15=4,p^2-4p-15=0 ?

The solutions to the equation 4p^2-4p-15=4,p^2-4p-15=0 are p=(1+2sqrt(5))/2 ,p=(1-2sqrt(5))/2

Find the zeros of 4p^2-4p-15=4,p^2-4p-15=0

The zeros of 4p^2-4p-15=4,p^2-4p-15=0 are p=(1+2sqrt(5))/2 ,p=(1-2sqrt(5))/2

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4 p^{2} - 4 p - 15 = 4, p^{2} - 4 p - 15 = 0

p = \frac{1 + 2 5}{2}, p = \frac{1 - 2 5}{2}

Decimal Notation

p = 2.73606 \dots, p = - 1.73606 \dots

Solution steps

4 p^{2} - 4 p - 15 = 4

Move

4

to the left side

4 p^{2} - 4 p - 19 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 ( - 19 )}{2 \cdot 4}

(- 4)^{2} - 4 \cdot 4 (- 19) = 85

p_{1, 2} = \frac{- ( - 4 ) \pm 8 5}{2 \cdot 4}

Separate the solutions

p_{1} = \frac{- ( - 4 ) + 8 5}{2 \cdot 4}, p_{2} = \frac{- ( - 4 ) - 8 5}{2 \cdot 4}

p = \frac{- ( - 4 ) + 8 5}{2 \cdot 4} : \frac{1 + 2 5}{2}

p = \frac{- ( - 4 ) - 8 5}{2 \cdot 4} : \frac{1 - 2 5}{2}

The solutions to the quadratic equation are:

p = \frac{1 + 2 5}{2}, p = \frac{1 - 2 5}{2}

0.5 (x^{2} + 1) = x

- x^{2} + x + 7 = 0

x^{2} + 12 x + 36 = 0

- y^{2} = 1

x, x^{2} + y^{2} + 20 x + 2 y + 102 = 0

What are the solutions to the equation 4p^2-4p-15=4,p^2-4p-15=0 ?
The solutions to the equation 4p^2-4p-15=4,p^2-4p-15=0 are p=(1+2sqrt(5))/2 ,p=(1-2sqrt(5))/2
Find the zeros of 4p^2-4p-15=4,p^2-4p-15=0
The zeros of 4p^2-4p-15=4,p^2-4p-15=0 are p=(1+2sqrt(5))/2 ,p=(1-2sqrt(5))/2