Question

7(2x²-x-3)+4x(2x+5)=

Answer 1

$7(2x^2 - x - 3) + 4x(2x + 5) \\ \\ 14x^2 - 7x - 21 + 8x^2 + 20x \\ \\ \\ => 22x^2 + 13 - 21 = 0$

Resolvendo por Bháskara:


$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=22, b=13, c=−21

Δ=b²−4ac
Δ=(13)2−4*(22)*(−21)
Δ=169+1848
Δ=2017


$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-13 \pm \sqrt{2017}}{2*22} \\ \\ \\ x = \dfrac{-13 \pm \sqrt{2017}}{44} \\ \\ \\ x' = \dfrac{-13 + \sqrt{2017}}{44} \\ \\ \\ \\ x'' = \dfrac{-13 - \sqrt{2017}}{44}$[/tex]


S = {$\dfrac{-13 + \sqrt{2017}}{44}, \ \ \dfrac{-13 - \sqrt{2017}}{44}$}