What are the solutions to the equation (2x+1)(x-4)=3 ?

The solutions to the equation (2x+1)(x-4)=3 are x=(7+sqrt(105))/4 ,x=(7-sqrt(105))/4

Find the zeros of (2x+1)(x-4)=3

The zeros of (2x+1)(x-4)=3 are x=(7+sqrt(105))/4 ,x=(7-sqrt(105))/4

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(2 x + 1) (x - 4) = 3

x = \frac{7 + 105}{4}, x = \frac{7 - 105}{4}

Decimal Notation

x = 4.31173 \dots, x = - 0.81173 \dots

Solution steps

(2 x + 1) (x - 4) = 3

Expand

(2 x + 1) (x - 4) : 2 x^{2} - 7 x - 4

2 x^{2} - 7 x - 4 = 3

Move

3

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 7) = 105

x_{1, 2} = \frac{- ( - 7 ) \pm 105}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 105}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 105}{2 \cdot 2}

x = \frac{- ( - 7 ) + 105}{2 \cdot 2} : \frac{7 + 105}{4}

x = \frac{- ( - 7 ) - 105}{2 \cdot 2} : \frac{7 - 105}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 105}{4}, x = \frac{7 - 105}{4}

3 a^{2} + 22 a - 45

\frac{3 x - 1}{∣ x - 1 ∣} \geq 4

14 x = 10

x^{6} + c^{2} x^{4} + c^{4} x^{2} + c^{6}

3 s^{4} t + 3 s^{3} t - 6 s^{2} t

What are the solutions to the equation (2x+1)(x-4)=3 ?
The solutions to the equation (2x+1)(x-4)=3 are x=(7+sqrt(105))/4 ,x=(7-sqrt(105))/4
Find the zeros of (2x+1)(x-4)=3
The zeros of (2x+1)(x-4)=3 are x=(7+sqrt(105))/4 ,x=(7-sqrt(105))/4