What are the solutions to the equation (2x+5)(x-4)=-5 ?

The solutions to the equation (2x+5)(x-4)=-5 are x=(3+sqrt(129))/4 ,x=(3-sqrt(129))/4

Find the zeros of (2x+5)(x-4)=-5

The zeros of (2x+5)(x-4)=-5 are x=(3+sqrt(129))/4 ,x=(3-sqrt(129))/4

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(2 x + 5) (x - 4) = - 5

x = \frac{3 + 129}{4}, x = \frac{3 - 129}{4}

Decimal Notation

x = 3.58945 \dots, x = - 2.08945 \dots

Solution steps

(2 x + 5) (x - 4) = - 5

Expand

(2 x + 5) (x - 4) : 2 x^{2} - 3 x - 20

2 x^{2} - 3 x - 20 = - 5

Move

5

to the left side

2 x^{2} - 3 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 15) = 129

x_{1, 2} = \frac{- ( - 3 ) \pm 129}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 129}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 129}{2 \cdot 2}

x = \frac{- ( - 3 ) + 129}{2 \cdot 2} : \frac{3 + 129}{4}

x = \frac{- ( - 3 ) - 129}{2 \cdot 2} : \frac{3 - 129}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 129}{4}, x = \frac{3 - 129}{4}

x^{2} + x = 30

\frac{5}{2} x + \frac{1}{2} x = 5 + \frac{7}{2} x

0^{- 3}

(x + 7)^{2} = 4

- \frac{7 π}{4} - 2 π

What are the solutions to the equation (2x+5)(x-4)=-5 ?
The solutions to the equation (2x+5)(x-4)=-5 are x=(3+sqrt(129))/4 ,x=(3-sqrt(129))/4
Find the zeros of (2x+5)(x-4)=-5
The zeros of (2x+5)(x-4)=-5 are x=(3+sqrt(129))/4 ,x=(3-sqrt(129))/4