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21-4x^2-5x=0
a = -4; b = -5; c = +21;
Δ = b2-4ac
Δ = -52-4·(-4)·21
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*-4}=\frac{-14}{-8} =1+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*-4}=\frac{24}{-8} =-3 $
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