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Kojom formulom tražimo rješenja kvadratne jednadžbe?
x1,2=b±b2−4ac2ax_{1,2}=\frac{b\pm\sqrt{b^2-4ac}}{2a}x1,2=2ab±b2−4ac
x1,2=−bb2−4ac2ax_{1,2}=\frac{-b\sqrt{b^2-4ac}}{2a}x1,2=2a−bb2−4ac
x1,2=−b±b2−4ac2ax_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}x1,2=2a−b±b2−4ac
x1,2=−b±b2−4ac4ax_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{4a}x1,2=4a−b±b2−4ac
ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0
Kako se zovu članovi kvadratne jednadžbe?
−2x2−5x+7=0-2x^2-5x+7=0−2x2−5x+7=0
Koja su rješenja ove kvadratne jednadžbe?
x1=−12, x2=−2x_1=\frac{-1}{2},\ x_2=-2x1=2−1, x2=−2
x1=−72, x2=1x_1=\frac{-7}{2},\ x_2=1x1=2−7, x2=1
x1=72, x2=−1x_1=\frac{7}{2},\ x_2=-1x1=27, x2=−1
x1=12, x2=2x_1=\frac{1}{2},\ x_2=2x1=21, x2=2
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