What are the solutions to the equation 2(2x^2+3)=11x ?

The solutions to the equation 2(2x^2+3)=11x are x=2,x= 3/4

Find the zeros of 2(2x^2+3)=11x

The zeros of 2(2x^2+3)=11x are x=2,x= 3/4

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2 (2 x^{2} + 3) = 11 x

x = 2, x = \frac{3}{4}

Decimal Notation

x = 2, x = 0.75

Solution steps

2 (2 x^{2} + 3) = 11 x

Expand

2 (2 x^{2} + 3) : 4 x^{2} + 6

4 x^{2} + 6 = 11 x

Move

11 x

to the left side

4 x^{2} + 6 - 11 x = 0

Write in the standard form

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

4 x^{2} - 11 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 \cdot 6}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 \cdot 6 = 5

x_{1, 2} = \frac{- ( - 11 ) \pm 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 5}{2 \cdot 4}, x_{2} = \frac{- ( - 11 ) - 5}{2 \cdot 4}

x = \frac{- ( - 11 ) + 5}{2 \cdot 4} : 2

x = \frac{- ( - 11 ) - 5}{2 \cdot 4} : \frac{3}{4}

The solutions to the quadratic equation are:

x = 2, x = \frac{3}{4}

(\frac{- 6 a ^{- 9} b ^{5}}{2 a ^{2} b ^{- 4}})^{4}

2 a^{2} - 9 a - 45

- \frac{1}{2} (2) + 1

2 x + 1 \leq - 5

(a^{2} b c^{3})^{4} \cdot (b^{2} c)^{3}

What are the solutions to the equation 2(2x^2+3)=11x ?
The solutions to the equation 2(2x^2+3)=11x are x=2,x= 3/4
Find the zeros of 2(2x^2+3)=11x
The zeros of 2(2x^2+3)=11x are x=2,x= 3/4