What are the solutions to the equation (x+3)^2+(x-6)(x+5)=0 ?

The solutions to the equation (x+3)^2+(x-6)(x+5)=0 are x=(-5+sqrt(193))/4 ,x=(-5-sqrt(193))/4

Find the zeros of (x+3)^2+(x-6)(x+5)=0

The zeros of (x+3)^2+(x-6)(x+5)=0 are x=(-5+sqrt(193))/4 ,x=(-5-sqrt(193))/4

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(x + 3)^{2} + (x - 6) (x + 5) = 0

x = \frac{- 5 + 193}{4}, x = \frac{- 5 - 193}{4}

Decimal Notation

x = 2.22311 \dots, x = - 4.72311 \dots

Solution steps

(x + 3)^{2} + (x - 6) (x + 5) = 0

Expand

(x + 3)^{2} + (x - 6) (x + 5) : 2 x^{2} + 5 x - 21

2 x^{2} + 5 x - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 21 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 21) = 193

x_{1, 2} = \frac{- 5 \pm 193}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 193}{2 \cdot 2}, x_{2} = \frac{- 5 - 193}{2 \cdot 2}

x = \frac{- 5 + 193}{2 \cdot 2} : \frac{- 5 + 193}{4}

x = \frac{- 5 - 193}{2 \cdot 2} : \frac{- 5 - 193}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 193}{4}, x = \frac{- 5 - 193}{4}

r^{2} + z^{2} = 9

25 t - 0.05 t^{2} = 15 t + 200

x^{2} + 1 \frac{2}{3} x - \frac{2}{3} = 0

7 x^{2} - 13 = 0

2 a^{2} + a - 23 = - 2

What are the solutions to the equation (x+3)^2+(x-6)(x+5)=0 ?
The solutions to the equation (x+3)^2+(x-6)(x+5)=0 are x=(-5+sqrt(193))/4 ,x=(-5-sqrt(193))/4
Find the zeros of (x+3)^2+(x-6)(x+5)=0
The zeros of (x+3)^2+(x-6)(x+5)=0 are x=(-5+sqrt(193))/4 ,x=(-5-sqrt(193))/4