What are the solutions to the equation (2y+5)(2y-5)+10y=0 ?

The solutions to the equation (2y+5)(2y-5)+10y=0 are y=(5(sqrt(5)-1))/4 ,y=-(5(1+sqrt(5)))/4

Find the zeros of (2y+5)(2y-5)+10y=0

The zeros of (2y+5)(2y-5)+10y=0 are y=(5(sqrt(5)-1))/4 ,y=-(5(1+sqrt(5)))/4

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(2 y + 5) (2 y - 5) + 10 y = 0

y = \frac{5 ( 5 - 1 )}{4}, y = - \frac{5 ( 1 + 5 )}{4}

Decimal Notation

y = 1.54508 \dots, y = - 4.04508 \dots

Solution steps

(2 y + 5) (2 y - 5) + 10 y = 0

Expand

(2 y + 5) (2 y - 5) + 10 y : 4 y^{2} - 25 + 10 y

4 y^{2} - 25 + 10 y = 0

Write in the standard form

a x^{2} + b x + c = 0

4 y^{2} + 10 y - 25 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 4 ( - 25 )}{2 \cdot 4}

1 0^{2} - 4 \cdot 4 (- 25) = 105

y_{1, 2} = \frac{- 10 \pm 10 5}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 10 + 10 5}{2 \cdot 4}, y_{2} = \frac{- 10 - 10 5}{2 \cdot 4}

y = \frac{- 10 + 10 5}{2 \cdot 4} : \frac{5 ( 5 - 1 )}{4}

y = \frac{- 10 - 10 5}{2 \cdot 4} : - \frac{5 ( 1 + 5 )}{4}

The solutions to the quadratic equation are:

y = \frac{5 ( 5 - 1 )}{4}, y = - \frac{5 ( 1 + 5 )}{4}

2 x^{2} + 5 x = 6 x + 15

x, 2 x^{2} + 4 x y - x - 2 y = 0

3 (a + 2)^{2} = 2 (a^{2} + 6 a - 1)

2 x^{2} = - 4 x - 1

3 y^{2} + 4 y - 32 = 0

What are the solutions to the equation (2y+5)(2y-5)+10y=0 ?
The solutions to the equation (2y+5)(2y-5)+10y=0 are y=(5(sqrt(5)-1))/4 ,y=-(5(1+sqrt(5)))/4
Find the zeros of (2y+5)(2y-5)+10y=0
The zeros of (2y+5)(2y-5)+10y=0 are y=(5(sqrt(5)-1))/4 ,y=-(5(1+sqrt(5)))/4