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26=4x+5x^2
We move all terms to the left:
26-(4x+5x^2)=0
We get rid of parentheses
-5x^2-4x+26=0
a = -5; b = -4; c = +26;
Δ = b2-4ac
Δ = -42-4·(-5)·26
Δ = 536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{134}}{2*-5}=\frac{4-2\sqrt{134}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{134}}{2*-5}=\frac{4+2\sqrt{134}}{-10} $
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