Question

4z^2-11z+6=0 ejersisios de segundo grado y^2+13y-48=0

Answer 1

$ax^{2} +bx+c=0, x_{1} =\frac{-b+\sqrt{b^{2}-4ac}}{2a},x_{2} =\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

$1) 4z^{2}-11z+6=0 \\z_{1} =\frac{-(-11)+\sqrt{(-11)^{2}-4.4.6}}{2.4}=\frac{11+\sqrt{25}}{8}=2\\ z_{2} =\frac{-(-11)-\sqrt{(-11)^{2}-4.4.6}}{2.4}=\frac{11-\sqrt{25}}{8}=\frac{3}{4}$

$2) y^{2}+13y-48=0\\ y_{1} =\frac{-13+\sqrt{13^{2}-4.1.(-48)}}{2.1}=\frac{-13+\sqrt{361}}{2}=3\\ y_{2} =\frac{-13-\sqrt{13^{2}-4.1.(-48)}}{2.1}=\frac{-13-\sqrt{361}}{2}=\frac{-27}{2}$