Solve -39+(2x-3)^2=2.(-10)

$$- 39 + ( 2 x - 3 ) ^ { 2 } = 2 . ( - 10 )$$

Solve for x

$$x = \frac{\sqrt{41} + 3}{2} \approx 4.701562119$$

$$x=\frac{3-\sqrt{41}}{2}\approx -1.701562119$$

Steps by Finding Square Root

Add 39 to both sides of the equation.

$$\left(2x-3\right)^{2}-39+39=2+39$$

Subtracting 39 from itself leaves 0.

$$\left(2x-3\right)^{2}=2+39$$

Add 2 to 39.

$$\left(2x-3\right)^{2}=41$$

Take the square root of both sides of the equation.

$$2x-3=\sqrt{41} 2x-3=-\sqrt{41}$$

Add 3 to both sides of the equation.

$$2x-3-\left(-3\right)=\sqrt{41}-\left(-3\right) 2x-3-\left(-3\right)=-\sqrt{41}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$2x=\sqrt{41}-\left(-3\right) 2x=-\sqrt{41}-\left(-3\right)$$

Subtract -3 from \sqrt{41}.

$$2x=\sqrt{41}+3$$

Subtract -3 from -\sqrt{41}.

$$2x=3-\sqrt{41}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\sqrt{41}+3}{2} \frac{2x}{2}=\frac{3-\sqrt{41}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\sqrt{41}+3}{2} x=\frac{3-\sqrt{41}}{2}$$