What are the solutions to the equation 4x^2-3=16x ?

The solutions to the equation 4x^2-3=16x are x=(4+sqrt(19))/2 ,x=(4-sqrt(19))/2

Find the zeros of 4x^2-3=16x

The zeros of 4x^2-3=16x are x=(4+sqrt(19))/2 ,x=(4-sqrt(19))/2

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4 x^{2} - 3 = 16 x

x = \frac{4 + 19}{2}, x = \frac{4 - 19}{2}

Decimal Notation

x = 4.17944 \dots, x = - 0.17944 \dots

Solution steps

4 x^{2} - 3 = 16 x

Move

16 x

to the left side

4 x^{2} - 3 - 16 x = 0

Write in the standard form

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

4 x^{2} - 16 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 (- 3) = 419

x_{1, 2} = \frac{- ( - 16 ) \pm 4 19}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4 19}{2 \cdot 4}, x_{2} = \frac{- ( - 16 ) - 4 19}{2 \cdot 4}

x = \frac{- ( - 16 ) + 4 19}{2 \cdot 4} : \frac{4 + 19}{2}

x = \frac{- ( - 16 ) - 4 19}{2 \cdot 4} : \frac{4 - 19}{2}

The solutions to the quadratic equation are:

x = \frac{4 + 19}{2}, x = \frac{4 - 19}{2}

2.25

5 (x - 1) + 2 x = - x + 7

12 x^{2} + x = 6

\frac{5 π}{12} \cdot \frac{180}{π}

\frac{3 m ^{4}}{3 m + 2} - \frac{4 m ^{2}}{5 m + 2}

What are the solutions to the equation 4x^2-3=16x ?
The solutions to the equation 4x^2-3=16x are x=(4+sqrt(19))/2 ,x=(4-sqrt(19))/2
Find the zeros of 4x^2-3=16x
The zeros of 4x^2-3=16x are x=(4+sqrt(19))/2 ,x=(4-sqrt(19))/2