What are the solutions to the equation x+x^2-25=15x+47 ?

The solutions to the equation x+x^2-25=15x+47 are x=18,x=-4

Find the zeros of x+x^2-25=15x+47

The zeros of x+x^2-25=15x+47 are x=18,x=-4

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x + x^{2} - 25 = 15 x + 47

x = 18, x = - 4

Solution steps

x + x^{2} - 25 = 15 x + 47

Move

47

to the left side

x^{2} + x - 72 = 15 x

Move

15 x

to the left side

x^{2} - 14 x - 72 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot ( - 72 )}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot (- 72) = 22

x_{1, 2} = \frac{- ( - 14 ) \pm 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 22}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 22}{2 \cdot 1}

x = \frac{- ( - 14 ) + 22}{2 \cdot 1} : 18

x = \frac{- ( - 14 ) - 22}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 18, x = - 4

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

x^{2} - 4 = 23

x^{2} - 80 x + 1200 = 0

x^{2} - 9 x + 12 = 0

a^{2} + 1 6^{2} = 3 4^{2}

What are the solutions to the equation x+x^2-25=15x+47 ?
The solutions to the equation x+x^2-25=15x+47 are x=18,x=-4
Find the zeros of x+x^2-25=15x+47
The zeros of x+x^2-25=15x+47 are x=18,x=-4