122)
$x^2+4x-1=0$
$a=1$;
$b=4$;
$c=-1$
$∆= b^2-4ac = 4^2-(4·1·-1) = 16-(-4) = 16+4 = 20$;
$x_{1,2}= \dfrac{-b±\sqrt{∆}}{2a} = \dfrac{-4±\sqrt{20}}{2·1} = \dfrac{-4±2\sqrt5}{2}$
quindi:
$x_1= \dfrac{-4-2\sqrt5}{2} = -2-\sqrt5~(≅-4,236068)$;
$x_2= \dfrac{-4+2\sqrt5}{2} = -2+\sqrt5~(≅+0,236068)$.
x^2+4x-1 = 0
x = (-4±√(-4)^2-(4*-1*1)
x = (-4±√20)/2
x = (-4±2√5)/2= -2±√5
x1 = -2+√5
x2 = -2-√5
verifica :
(-2+√5)^2+4x = 1
4+5-4√5-8+4√5 = 1
9-8 = 1
(-2-√5)^2+4x = 1
4+5+4√5-8-4√5 = 1
9-8 = 1 ....funziona 👍😊