Answer:
x₁ = ⁵/₄ , x₂ = 0
Step-by-step explanation:
[tex]13x^2-16x=x^2-x\\\\13x^2-16x-x^2+x=0\\\\12x^2-15x=0\\{\qquad}^{\div3}\qquad\quad^{\div3}\\4x^2-5x=0\quad\implies\quad a=4\,,\ b=-5\,,\ c=0\\\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\cdot4\cdot0}}{2\cdot4}=\dfrac{5\pm\sqrt{25}}8\\\\\\x_1=\dfrac{5+5}8=\dfrac54\,,\qquad x_2=\dfrac{5-5}8=0[/tex]
