Question

Anyone please help me​

Answer 1

$\\ \sf\longmapsto x^2-4x-5=0$

• Use midterm splitation

$\\ \sf\longmapsto x^2-5x+x-5=0$

$\\ \sf\longmapsto x(x-5)+1(x-5)=0$

$\\ \sf\longmapsto (x-5)(x+1)=0$

$\\ \sf\longmapsto x=5\:or\:x=-1$

Solution is (-1,0) and (5,0)

Graph attached.

Spare way:-

Use quadratic formula .

$\\ \sf\longmapsto x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\\ \sf\longmapsto x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}$

$\\ \sf\longmapsto x=\dfrac{4\pm \sqrt{16+20}}{2}$

$\\ \sf\longmapsto x=\dfrac{4\pm 6}{2}$

$\\ \sf\longmapsto x=\dfrac{10}{2}\:or\:\dfrac{-2}{2}$

$\\ \sf\longmapsto x=5\:or\:-1$

Answer 2

$x {}^{2} - 4x - 5 = 0 \\ x {}^{2} + x - 5x - 5 = 0 \\ x(x + 1) - 5(x + 1) = 0 \\ (x + 1)(x - 5) = 0 \\ x + 1 = 0 \\ x - 5 = 0 \\ x = - 1 \\ x = 5$

• The equation has two solutions:

x=5, x=-1

Graph is in the attachment!!