What are the solutions to the equation -4y^2-12y+5=0 ?

The solutions to the equation -4y^2-12y+5=0 are y=-(3+sqrt(14))/2 ,y=(sqrt(14)-3)/2

Find the zeros of -4y^2-12y+5=0

The zeros of -4y^2-12y+5=0 are y=-(3+sqrt(14))/2 ,y=(sqrt(14)-3)/2

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- 4 y^{2} - 12 y + 5 = 0

y = - \frac{3 + 14}{2}, y = \frac{14 - 3}{2}

Decimal Notation

y = - 3.37082 \dots, y = 0.37082 \dots

Solution steps

- 4 y^{2} - 12 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 12)^{2} - 4 (- 4) \cdot 5 = 414

y_{1, 2} = \frac{- ( - 12 ) \pm 4 14}{2 ( - 4 )}

Separate the solutions

y_{1} = \frac{- ( - 12 ) + 4 14}{2 ( - 4 )}, y_{2} = \frac{- ( - 12 ) - 4 14}{2 ( - 4 )}

y = \frac{- ( - 12 ) + 4 14}{2 ( - 4 )} : - \frac{3 + 14}{2}

y = \frac{- ( - 12 ) - 4 14}{2 ( - 4 )} : \frac{14 - 3}{2}

The solutions to the quadratic equation are:

y = - \frac{3 + 14}{2}, y = \frac{14 - 3}{2}

- \frac{1}{2} = \frac{3}{8} y

\frac{x ^{2} - 9}{x + 3}

22 = - 11 x

9 v^{2} - 48 v + 64

59 x^{2} - 122 = 0

What are the solutions to the equation -4y^2-12y+5=0 ?
The solutions to the equation -4y^2-12y+5=0 are y=-(3+sqrt(14))/2 ,y=(sqrt(14)-3)/2
Find the zeros of -4y^2-12y+5=0
The zeros of -4y^2-12y+5=0 are y=-(3+sqrt(14))/2 ,y=(sqrt(14)-3)/2