What are the solutions to the equation (x+7)(x-10)=1 ?

The solutions to the equation (x+7)(x-10)=1 are x=(3+sqrt(293))/2 ,x=(3-sqrt(293))/2

Find the zeros of (x+7)(x-10)=1

The zeros of (x+7)(x-10)=1 are x=(3+sqrt(293))/2 ,x=(3-sqrt(293))/2

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(x + 7) (x - 10) = 1

x = \frac{3 + 293}{2}, x = \frac{3 - 293}{2}

Decimal Notation

x = 10.05862 \dots, x = - 7.05862 \dots

Solution steps

(x + 7) (x - 10) = 1

Expand

(x + 7) (x - 10) : x^{2} - 3 x - 70

x^{2} - 3 x - 70 = 1

Move

1

to the left side

x^{2} - 3 x - 71 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 71 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 71) = 293

x_{1, 2} = \frac{- ( - 3 ) \pm 293}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 293}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 293}{2 \cdot 1}

x = \frac{- ( - 3 ) + 293}{2 \cdot 1} : \frac{3 + 293}{2}

x = \frac{- ( - 3 ) - 293}{2 \cdot 1} : \frac{3 - 293}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 293}{2}, x = \frac{3 - 293}{2}

(1 - 7 i)^{2}

∣ 2 x - 3 ∣ - 4 \leq 5

∣ - 4 + 7 ∣

\frac{2}{5} - \frac{1}{10}

- 18 x + 2 x^{2} - 72

What are the solutions to the equation (x+7)(x-10)=1 ?
The solutions to the equation (x+7)(x-10)=1 are x=(3+sqrt(293))/2 ,x=(3-sqrt(293))/2
Find the zeros of (x+7)(x-10)=1
The zeros of (x+7)(x-10)=1 are x=(3+sqrt(293))/2 ,x=(3-sqrt(293))/2