What are the solutions to the equation 2x^2+3x-7=x^2+5x+39 ?

The solutions to the equation 2x^2+3x-7=x^2+5x+39 are x=1+sqrt(47),x=1-sqrt(47)

Find the zeros of 2x^2+3x-7=x^2+5x+39

The zeros of 2x^2+3x-7=x^2+5x+39 are x=1+sqrt(47),x=1-sqrt(47)

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2 x^{2} + 3 x - 7 = x^{2} + 5 x + 39

x = 1 + 47, x = 1 - 47

Decimal Notation

x = 7.85565 \dots, x = - 5.85565 \dots

Solution steps

2 x^{2} + 3 x - 7 = x^{2} + 5 x + 39

Move

39

to the left side

2 x^{2} + 3 x - 46 = x^{2} + 5 x

Move

5 x

to the left side

2 x^{2} - 2 x - 46 = x^{2}

Move

x^{2}

to the left side

x^{2} - 2 x - 46 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 46 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 46) = 247

x_{1, 2} = \frac{- ( - 2 ) \pm 2 47}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 47}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 47}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 47}{2 \cdot 1} : 1 + 47

x = \frac{- ( - 2 ) - 2 47}{2 \cdot 1} : 1 - 47

The solutions to the quadratic equation are:

x = 1 + 47, x = 1 - 47

\frac{\frac{3}{8} - \frac{3}{4}}{\frac{2}{3} - \frac{3}{5}}

(x + 3)^{2} = - 49

3 c^{3} - 6 c

x^{2} - 5 x + 6 \leq 0

\frac{180}{π} \cdot \frac{5 π}{6}

What are the solutions to the equation 2x^2+3x-7=x^2+5x+39 ?
The solutions to the equation 2x^2+3x-7=x^2+5x+39 are x=1+sqrt(47),x=1-sqrt(47)
Find the zeros of 2x^2+3x-7=x^2+5x+39
The zeros of 2x^2+3x-7=x^2+5x+39 are x=1+sqrt(47),x=1-sqrt(47)